Problem: What is the area of the region between the graphs of $f(x)=x^2-3x$ and $g(x)=2x$ from $x=0$ to $x=5$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{325}{6}$ (Choice B) B $25$ (Choice C) C $\dfrac{175}{6}$ (Choice D) D $\dfrac{125}{6}$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between $x=0$ and $x=5$. From this we are looking to evaluate: $ \int_{0}^{5}\left( g(x)-f(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{0}^{5} \left( 2x- (x^2-3x) \right) \,dx \\\\ &= \int_{0}^{5} \left( 5x- x^2 \right) \,dx\\\\ &=\dfrac{5x^2}{2}-\dfrac{x^3}{3}~\Bigg|_{0}^{5} \\\\ &= \left( \dfrac{125}{2}-\dfrac{125}{3}\right) -\left( 0-0\right)\\\\ &= \dfrac{125}{6} \end{aligned}$ Answer The area is $\dfrac{125}{6}$ square units.